A race car is one lap behind the lead race car when the lead car has 57 laps to go in a race. If the speed of the lead car is 66.2 m/s, what must be the average speed of the second car to catch the lead car just before the end of the race (i.e., right at the finish line)? Assume 1 lap is 1.34 km.
Answer in units of m/s
2 answers
Never mind I figured it out the answer is 67.3614 if anyone needs help
How long before the lead car finishes?
57 * 1.34 = 76.38 km = 76.38*10^3 m
t = distance/speed = 76.38*10^3/66.2 = 1157 seconds
So in 1157 seconds second car must go
58 * 1.34 = 77.72 km = 77.72*10^3 m
speed = distance/time = 77.72/1.157 = 67.2 m/s
57 * 1.34 = 76.38 km = 76.38*10^3 m
t = distance/speed = 76.38*10^3/66.2 = 1157 seconds
So in 1157 seconds second car must go
58 * 1.34 = 77.72 km = 77.72*10^3 m
speed = distance/time = 77.72/1.157 = 67.2 m/s
1 answer